The electric field required to keep a water drop of mass 'm' just to remain suspended in air, when charged with one electron, is

An oil drop, carrying six electronic charges and having a mass of 1.6xx10^(-12)g , falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upward with the same speed as it was formerly moving down ward with ? Ignore buoyancy.

The electric charge required for electrode deposition of one gram-equivalent of a substance is :

Calculate the electric field strength required just to support a water drop of mass 10^(-7)kg having a charge 1.6 xx 10^(-19) C

Calculate the electric field strength which is required to just support a water drop of mass 10^(-3) kg and having a charge 1.6xx10^(-19)C .

An oil drop having a charge q and apparent weight W falls with a terminal speed v in absence of electric field . When electric field is switched on it starts moving up with terminal speed v. The strength of electric field is

A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100V/m makes it to move in a straight path, then the mass of the particle is ( Given charge of electron = 1.6 xx 10^(-19)C )

Mention the charge, mass and e/m ratio for an electron.

A charged oil drop falls with terminal velocity v_(0) in the absence of electric field. An electric field E keeps it stationary. The drop acquires charge 3q, it starts moving upwards with velocity v_(0) . The initial charge on the drop is

A point charge q is rotated along a circle in the electric field generated by anotherj point charge Q. The work done by the electric field on the rotatin charge in one complete revolution is

The electric field strength in N C^(-1) that is required to just prevent a water drop carrying a change 16 xx 10^(-19) C from falling under gravity is (g = 9.8 m s^(-2) , the mass of water drop = 0.0016 g )